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Author Topic: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011  (Read 670063 times)

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1410 on: May 30, 2011, 01:00:00 AM »
lol  How does one get a wattage value without the computation of current?

The value of the CSR is assumed to be approximately 1 Ohm for all current calculations. It will soon be shown what the correct value is however.

Regardless of the value of the CSR (anywhere from 0.25 Ohms to 2 Ohms for eg.), the trend being shown will be the same, i.e. the net average power declining in negative amplitude, and heading towards a positive value.

.99

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1411 on: May 30, 2011, 01:12:50 AM »
Continuing with the battery voltage probe placement closer and closer to the battery array, the results continue to show a declining negative average power in the battery array.

The first test run in this installment is with the voltage probes placed across the battery array and CSR (V1-3), completely eliminating the long battery wire leads. The power computation comes to -20W.

The second test run is with the battery voltage probes across nodes (V1-2), which eliminates the voltage across the CSR, and is therefore directly across the battery array and the associated battery jumper wires. This power computation comes to -17.5W.

More to follow.

.99

Still with this question here.  How does one determine the current flow through the battery without computing the value of current?  Please note.  You have stated that you managed this without the use of the CSR.  What did you use to determine current flow?  If you want us to assume that this is a serious exercise then you need to explain how you managed this calculation.  And correctly you need to show us precise points on a schematic and show that schematic here together with the waveforms and the math traces - as you did the others. 

And you've now obviated ALL reference to instantaneous vi dt which is the second point of the report.  We do NOT need a negative voltage across the shunt to prove a negative wattage as indicated in the negative math trace referenced repeatedly.  I think you need to LOSE that AVERAGING that you are now relying on.

Regards
Rosemary

Added. 
And more added.  This is what I'm referring to.  Your quote 'which eliminates the voltage across the CSR, and is therefore directly across the battery array and the associated battery jumper wires.'
« Last Edit: May 30, 2011, 01:33:01 AM by Rosemary Ainslie »

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1412 on: May 30, 2011, 06:13:38 AM »
How does one determine the current flow through the battery without computing the value of current?
The instantaneous current i(t) through the battery is determined by the instantaneous voltage v(t) across the CSR (denoted by "VCSR" in all the scope shots), divided by the CSR resistance.

If we use 1 Ohm for the CSR resistance, the instantaneous current i(t) is equal to the instantaneous voltage v(t) across the CSR. For the sake of illustration, I have used (and noted) 1 Ohm for the CSR value for most of my computations.

Therefore, the instantaneous and average power results are based on the instantaneous voltage across the CSR (which represents battery current directly), times the instantaneous voltage across the battery array (at various points as noted on the schematics and text throughout).

.99

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1413 on: May 30, 2011, 09:17:42 AM »
Hello Poynty
You seem to be implying that the voltage values vary depending on where you position the probes - as stated hereunder.

Continuing with the battery voltage probe placement closer and closer to the battery array, the results continue to show a declining negative average power in the battery array.

This would be expected.  What you are NOT doing is showing us precisely what changes you have made to your probe positions.  We need a schematic showing this.  We also need to see the same graphic waveforms and math trace values.  Please provide this.  Else there is no knowing what you are doing.

Then you state that you completely eliminate any copper between the CSR and the battery - hereunder.  Not sure how we can manage this without interconnecting leads - nor even why we should bother if this is giving us a benefit.  In any event ...

The first test run in this installment is with the voltage probes placed across the battery array and CSR (V1-3), completely eliminating the long battery wire leads. The power computation comes to -20W.

Now you get to a computation of -20 Watts - which, incidentally, is still in excess of our own results.  We're looking for a value closer to -7 watts when we have a zero discharge from the battery.  This is in line with Test 1 of our report.

Then you go on to say the following - which makes no sense at all.

The second test run is with the battery voltage probes across nodes (V1-2), which eliminates the voltage across the CSR, and is therefore directly across the battery array and the associated battery jumper wires. This power computation comes to -17.5W.

The puzzle here is that there is absolutely NO WAY you can evaluate any power delivered by or returned to the battery if you are NOT basing this on a wattage computation.  And for this particular computation you actually REQUIRE the evaluation of current.  And for this, AGAIN, you NEED the CSR factored into that product or you are not applying vi dt.  Nor are you showing us the variation to the oscillations and the math trace that we have rather grown to rely on, for this computation.  Did you make a mistake?  Or did you somehow run the PSpice program without reference to current?  Or did you simply ASSUME a 1 Ohm value to the shunt resistor and factor this into your analysis?  You see Poynty?  We are now dealing with allegations of yours and - dare I say it - implications.  This is hardly scientific.  It certainly is NOT good reporting. And what you are implying - or in fact stating - is that the there is a steadily reduced amount of energy being returned to the battery as one eliminates the inductance on the circuit.  Frankly - that's hardly surprising.  It's certainly consistent with our own findings.  BUT.  We have NO idea if you are even getting that oscillation on these new results.  Nor do we know it's amplitude nor its frequency.  All this was presented before - in clear schematics and downloads that were also VERY READABLE. For some reason you are now hiding this information.  And, consequently we now know nothing except what you are trying very hard to IMPLY.

Then you tell us - rather enticingly - 'more to follow'.  But what we actually got is only this.

The instantaneous current i(t) through the battery is determined by the instantaneous voltage v(t) across the CSR (denoted by "VCSR" in all the scope shots), divided by the CSR resistance.

I trust that you're not presuming to educate the most our readers here because I'm reasonably satisfied that we all know a little about elementary power analysis.  But what is confusing is that you now state that you have IN FACT factored in the current flow based on the CSR's resistance.  So.  How then do you justify your previous denial of this?  Here is that statement again.

'The second test run is with the battery voltage probes across nodes (V1-2), which eliminates the voltage across the CSR, and is therefore directly across the battery array and the associated battery jumper wires.'

Was this an error?  Did you not in fact mean this?  Which leaves the most of us wondering WHAT you IN FACT meant.  Please advise.

Anyway.  Moving on.  You then write ....

If we use 1 Ohm for the CSR resistance, the instantaneous current i(t) is equal to the instantaneous voltage v(t) across the CSR. For the sake of illustration, I have used (and noted) 1 Ohm for the CSR value for most of my computations.

Does this mean that you have ASSUMED a value for CSR?  And since it's compatible with your previous computation - then can we ASSUME that you are factoring in the required impedance because that oscillation is still there?  You see Poynty Point?  We no longer know what you're referring  to because you are NOT showing us the values directly off your simulation program.  And - in any event - you have either computed current correctly or you've computed the current incorrectly - and we will NEVER KNOW.  Again.  This speaks to a certain want in accurate reporting - with respect.  It would give you the license to say what you liked and claim what you like.  We have supported our own evidence with copious screen downloads.  May we impose on you to do the same?  That way we can compare applies with apples.  And it would then lend a certain credibility to your reports of a 'diminishing' benefit - which is otherwise lacking.

And as for this following statement ...

Therefore, the instantaneous and average power results are based on the instantaneous voltage across the CSR (which represents battery current directly), times the instantaneous voltage across the battery array (at various points as noted on the schematics and text throughout).

Very confusing.   :o

The instantaneous power would be based on vi dt.  Average voltage across the shunt and the battery has NOTHING WHATSOEVER to do with vi dt.  I seem to recall that you averaged the power but NOT the actual voltage values.  Please advise.  Because if you HAVE averaged the voltages across the battery and the load then you have factored OUT the benefit of the 180 degree anti phase relationship of both battery and shunt voltages that ADDS to the general benefit.

However, the good news is that you're now getting towards the values that we ourselves compute during the oscillation phase of each duty cycle.  Still WAY too much measured benefit on your side.  But hopefully you'll get this to more realistic levels in due course.   Meanwhile - may we all impose on you to show us those waveforms.  I would have thought that screen downloads would be relatively easily enabled with PSpice.  I am not sure why you need to hide this evidence.  I know that even I can manage those downloads and I'm USELESS on this internet thing.  I'm sure your own skills are more than equal to it.  And, as it's quick and easy - I wonder if we can impose on you to do the same?  That way we can ourselves verify your claims and implications here.

Kind regards,
Rosemary

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1414 on: May 30, 2011, 09:29:18 AM »
Rose,

Before I attempt to answer any of your questions, I must ask you this:

Have you been downloading, opening, and viewing the pdf document files I've attached to my posts?

In case you have not, these pdf files (which should be readable on MACs) contain all the schematics and scope shots I have referred to.

.99

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1415 on: May 30, 2011, 09:34:35 AM »
Rose,

Before I attempt to answer any of your questions, I must ask you this:

Have you been downloading, opening, and viewing the pdf document files I've attached to my posts?

In case you have not, these pdf files (which should be readable on MACs) contain all the schematics and scope shots I have referred to.

.99

Ok.  If this is the problem - then the answer is NO.  I am entirely unable to open those files.  I get a short list of what looks like computer code.  But it's barely 8 lines.  But this is important Poynty.  Just keep the records easily accessible to ALL readers - please. 

And thanks for undertaking to answer those questions.  If they're answered in clear downloads - then all's well.  Just give us those downloads.

Ta muchly,
Rosie

Added
BTW - please see to it that the sizing is compatible with this thread page - else we'll get that fiasco that Glen managed that rendered EVERYTHING unreadable.

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1416 on: May 30, 2011, 09:52:03 AM »
Rose,

You can download and install a pdf reader for your MAC from here:

http://www.adobe.com/support/downloads/detail.jsp?ftpID=3987

You should have a pdf reader anyway Rose, otherwise how can you read the many documents posted on the forums and all over the internet?

In order for the schematic to be clear and readable (it is too long for me to screen capture) , I must be place it in the pdf file. This also keeps all the scope shots and schematics in one convenient place.

.99

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1417 on: May 30, 2011, 09:58:51 AM »
Rose,

You can download and install a pdf reader for your MAC from here:

http://www.adobe.com/support/downloads/detail.jsp?ftpID=3987

You should have a pdf reader anyway Rose, otherwise how can you read the many documents posted on the forums and all over the internet?

In order for the schematic to be clear and readable (it is too long for me to screen capture) , I must be place it in the pdf file. This also keeps all the scope shots and schematics in one convenient place.

.99

How about a compromise?  A schematic showing the variations - and a single shot of the waveforms and a single shot of the power analysis.  I CANNOT open those files.  I can indeed open most others.  And three downloads does NOT take that much time.  Surely?  That way you keep your own files pristine - but those of us who are following your argument can, at least, reference the salient proofs.  You will notice how seldom your files are downloaded.  Surely you see the problem here?

Thanks
Rosie

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1418 on: May 30, 2011, 07:41:15 PM »
Please refer to the previous posts for explanations of the schematics and scope shots.

.99

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1419 on: May 30, 2011, 07:46:12 PM »
And the other 3 sets, which brings this up to date with the "detailed_analysis04.pdf" document.

.99

Offline TinselKoala

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1420 on: May 30, 2011, 10:12:29 PM »
 8)

Offline TinselKoala

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1421 on: May 30, 2011, 10:51:55 PM »
 :'(

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1422 on: May 31, 2011, 07:51:11 AM »
8)
:'(

lol  Nice to see you following the JouleSeeker's good example by keeping things terse and technical. 

Poynty - I have included your schematics and results in my report.  Thanks for your permission.  I'll comment on your work when you've finally managed that zero benefit which I rather suspect is where you're trending.  You've got a couple of serious contradictions that need addressing.  But in the meantime I take it that your reference to not computing any voltage across the CSR was stated  in error.

Manwhile guys - I'm busy for the next few days circulating the circular.  That's assuming that both this computer and I survive any more interventions.  We're both suffering a surfeit of those 'delusions' that Cat refers to.

Kind regards,
Rosie
 :o ::)
 ;D

BTW - Here's the content of that circular - still subject to minor editing requirements.
http://newlightondarkenergy.blogspot.com/2011/05/123-cover-letter-to-be-circulated-with.html

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1423 on: May 31, 2011, 10:00:03 AM »
You've got a couple of serious contradictions that need addressing.
There are a couple of things that will become clearer with the next installment, but not sure I would call them contradictions.

Quote
But in the meantime I take it that your reference to not computing any voltage across the CSR was stated  in error.
I have not stated nor implied that, but apparently you've interpreted things that way somehow.

The voltage across the CSR is measured and shown. The current is computed using a CSR value of "1", therefore no computation is required; it is a direct measurement of current. The actual magnitude of the current and power is important, but not so critical in this exercise. We know the CSR value is somewhere between 0.25 Ohms and perhaps 2 Ohms. The value of the resistive part of the CSR is simply a multiplying factor and is not involved in the polarity of the power in question.

.99

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #1424 on: June 03, 2011, 03:17:37 AM »
Sorry it's been a bit longer for this next post, but I just returned from vacation on Tuesday afternoon, and I've needed a bit of recovery time. Back at work Wednesday too.  ::)



For the next installment of simulation test runs, it's necessary to establish some simple background theory:

If each of the 6 twelve-volt batteries in the battery array have approximately the same state of charge, terminal voltage, and internal resistance, it is reasonable to asssume that each of the 6 batteries will receive or supply the same amount of power in the circuit. As such, it is valid to measure and analyse the power in any one of the 6 batteries and apply a factor of 6x to obtain the total power in the circuit.

In this first test, the battery voltage probes are placed across the last jumper wire and last 12V battery. So we are measuring the voltage across a single 12V battery in series with 400nH of wire inductance in a single jumper. The power computes to -3.8W.

Next, when the battery voltage probes are placed directly across the single 12V battery and no jumper, the power changes polarity and computes to roughly +1.4W.

When the wattage probe available in PSpice is used to directly measure the instantaneous power of the single 12V battery, it computes to a net average of approximately -5.45W. If you recall the exercise on the polarity of power sources vs. power dissipators a little while back, you will know that the proper polarity for a source that is sourcing power, is negative. The reason the last computation of +1.4W turned out positive, is because the voltage probes across the CSR are reversed (as a matter of establishing common ground for both the CSR and battery probes). This has been the case throughout this exercise. It adds a bit of confusion, but that is the direction the "powers" normally go and it's important to keep this straight in one's mind.

Now back to the issue of the correct value for the CSR. As we now know the true power in any one of the six 12V batteries is about -5.45W, and that the previous measurement using a single 12V battery times the CSR voltage (battery current) came to approximately +1.4W (assuming a 1 Ohm value for the CSR), it may become obvious that assuming the CSR value to be anything other than 0.25 Ohms is incorrect. If we take the +1.4W measurement and multiply it by 4x (1/0.25), we obtain a power of about +5.6W. I have been approximating the values read off the scope, so in reality the previous measurement would actually be closer to +1.37W. It should be clear from this that the correct value for the CSR when looking at DC INPUT power, is the actual resistive value of the CSR, in this case 0.25 Ohms (regardless if the current is pulsed at a high frequency or not).

Computing the total power (using the Wattage probe) from all 6 batteries in the array we have:

-5.45W x 6 = -32.7W

This is the actual correct value and polarity for the total INPUT power of the battery array in this particular simulation.

Now, if we take the previous +1.37W measurement (which used the VCSR(t) x VBAT(t)) using just a single battery and no jumper wire, and multiply it by 4 (because of the 0.25 Ohm CSR), then by 6 (for 6 batteries in the array), we obtain a power of about +32.88W.

Other than the polarity difference (because the CSR probes are reversed), the two powers are almost identical in magnitude, and it is safe to say that now with the inductance eliminated in the battery voltage measurement, the VCSR(t) x VBAT(t) computation by the scope is very accurate.

More to follow.

.99
« Last Edit: June 03, 2011, 03:39:53 AM by poynt99 »